Problem: The equation of hyperbola $H$ is $\dfrac {(x+3)^{2}}{36}-\dfrac {(y-8)^{2}}{64} = 1$. What are the asymptotes?
We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y-8)^{2}}{64} = - 1 + \dfrac {(x+3)^{2}}{36}$ Multiply both sides of the equation by $64$ $(y-8)^{2} = { - 64 + \dfrac{ (x+3)^{2} \cdot 64 }{36}}$ Take the square root of both sides. $\sqrt{(y-8)^{2}} = \pm \sqrt { - 64 + \dfrac{ (x+3)^{2} \cdot 64 }{36}}$ $ y - 8 = \pm \sqrt { - 64 + \dfrac{ (x+3)^{2} \cdot 64 }{36}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y - 8 \approx \pm \sqrt {\dfrac{ (x+3)^{2} \cdot 64 }{36}}$ $y - 8 \approx \pm \left(\dfrac{8 \cdot (x + 3)}{6}\right)$ Add $8$ to both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{4}{3}(x + 3)+ 8$